Question:
If the line, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane,
$x+2 y+3 z=15$ at a point $P$, then the distance of $P$ from the origin is
Correct Option: 1
Solution:
Any point on the given line can be
$(1+2 \lambda,-1+3 \lambda, 2+4 \lambda) ; \lambda \in \mathrm{R}$
Put in plane
$1+2 \lambda+(-2+6 \lambda)+(6+12 \lambda)=15$
$20 \lambda+5=15$
$20 \lambda=10$
$\lambda=\frac{1}{2}$
$\therefore$ Point $\left(2, \frac{1}{2}, 4\right)$
Distance from origin
$=\sqrt{4+\frac{1}{4}+16}=\frac{\sqrt{16+1+64}}{2}=\frac{\sqrt{81}}{2}$
$=\frac{9}{2}$
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